dropping wall wart voltage to 9vdc

[runmikeyrun]

I'm not sure of the best way to do this- resistor, voltage divider or 9v voltage regulator?  I have multiple pedals on battery and multiple power supplies, but they are 12v (putting out a measured 15v).  One of them is about 1 amp so i'd like to power a few pedals with it. 

I'm thinking a 9v regulator setup would be best, but not sure how to build something like that.  Want to keep it simple so i can fit the whole thing into one of those super small hammond boxes.  Can i just run the 15v into the 9v regulator and put the output on the jacks?  Something possibly like this (courtesy of GGG)

 

[R.G.]

I'm not sure of the best way to do this- resistor, voltage divider or 9v voltage regulator?

Unless you have a fixed, unvarying load, a regulator is always the better way to keep a fixed voltage.

I'm thinking a 9v regulator setup would be best, but not sure how to build something like that.  Want to keep it simple so i can fit the whole thing into one of those super small hammond boxes.  Can i just run the 15v into the 9v regulator and put the output on the jacks?  Something possibly like this (courtesy of GGG)

That works fine as long as you don't cause the 7809 to get too hot by pulling too much current. Connect the tab of the 7809 to the metal box with some heat sink goo between them to transfer heat and you'll be able to provide more current. The 7809 shuts down if it gets too hot.

[runmikeyrun]

Thanks RG!  I will do that.  I'll check to see what the current limits are on 7809s at my local shop.  I actually have a couple heatsinks left over from scavenging computers for wire, but in that tiny of an enclosure it might not do that good once the air temp inside hits 150 degrees 

[Zben3129]

Semirelated question, I have a setup now with a 12 volt regulator, and I would like to get that 12v dc down to 9v dc with resistors if possible. Couldn't I just put a 3m9 and a 1m2 resistor in series from 12v to ground and tap 9v from where they meet? With a 100uf capacitor across 9v to ground?

Zach

[cpm]

you can save space for the smallish box by using only one output jack and daisy chaining

Semirelated question, I have a setup now with a 12 volt regulator, and I would like to get that 12v dc down to 9v dc with resistors if possible. Couldn't I just put a 3m9 and a 1m2 resistor in series from 12v to ground and tap 9v from where they meet? With a 100uf capacitor across 9v to ground?

i think those resistors wont drive more than some uA

[Zben3129]

thats what I was afraid of. I did some calculations and it looks like the voltage would sag to...well...nothing pretty much 


I am wondering if this is even feasable with resisors.


Zach

[R.G.]

Semirelated question, I have a setup now with a 12 volt regulator, and I would like to get that 12v dc down to 9v dc with resistors if possible. Couldn't I just put a 3m9 and a 1m2 resistor in series from 12v to ground and tap 9v from where they meet? With a 100uf capacitor across 9v to ground?

I typed in a whole long polemic on resistor networks here a few days? Weeks? Months? ago.

You can make a resistor divider provide any voltage lower than the supply voltage to it. However, the resulting voltage looks like it has a resistance in series with it equal to the parallel combination of the two resistors making the voltage. In your example, the resulting voltage is
V = 12* 3.9/(1.2+3.9) = 9.176V. But it looks like it has R = (3M9*1M2)/(3M9+1M2) = 918K ohms in series with it.

If you use 3.9K and 1.2K, then it looks like it has 918 ohms in series with it. In all cases, it sags the instant you put a load on it.

Resistor dividers only give you knowable voltages if the current drawn from them is negligible or if you know and can calculate in the effect of the loads.

[Zben3129]

Not sure if you were referring to the one you typed to me about voltage divider networks 

The difference between the two is the first one you know 2 variables and need to find the third. R=V/I, you know V and I. Simple enough.

The new problem involves finding 2 variables. V=IR. Solve for I and R (oh and don't forget that when I varies R must remain constant). Good luck on that impossible one, not even sure it could be called math 

Basically there is no way to make a power supply with resistors, unless it is speciallized. Not what I am looking for.

Other ways to drop 3 volts without a regulator that will remain constant V with varying I? 5 silicon diodes in a row? Sounds like it would work but somewhat inefficient.


Zach

[R.G.]

Basically there is no way to make a power supply with resistors, unless it is speciallized. Not what I am looking for.

There is no way to make a *constant voltage* power supply with resistors unless it has a constant load. If the load is constant, you only need one resistor.

Other ways to drop 3 volts without a regulator that will remain constant V with varying I? 5 silicon diodes in a row? Sounds like it would work but somewhat inefficient.

Depends on what you mean by "inefficient". If your input voltage is constant, and you want a constant output voltage for a varying current, then you can use any constant-voltage-drop device to get rid of the voltage. It doesn't matter what constant voltage device you use, it will always dissipate the same power - the voltage across it times the current through it; in that sense, all constant-voltage-drop devices for this application are equally efficient - they lose exactly the same heat.

If, however, your input voltage varies as well as the output current varying, there is nothing for it but to use an active regulator circuit. You can use either a series regulator circuit, the conceptual equal of the 780x regulators. You can also use a series resistor and a shunt zener equal to the desired output voltage. The zener sets the output voltage, and the series resistor sets the total load plus zener current. The load current can then vary from zero to the total current the resistor allows without changing the voltage.

[aron]

I've always used voltage regulators for this. It's simple and it works.

[Zben3129]

I've always used voltage regulators for this. It's simple and it works.

I could easily just put in the 9v regulator, which I will eventually do, but right now I am just trying to figure out alternate ways to do this/general voltage dropping. Or find out that its not really plausible, either way 

Thanks

Zach

[brett]

Hi
depending on the current capacity of that 12V supply, there are a few options, assuming that you want to ignore Aron's advice and do it the difficult and inefficient way instead of the simple and effective way.

Assuming that you've got 100mA or thereabout of current capacity, try putting a 22 ohm 5 W resistor to the +, in series with a 100 ohm 5W pot to ground.  Connect your rig to the junction.  Then adjust the pot until you get about 9V at the junction. As RG said, the voltage will sag with extra load and rise with less load.  But it will kind of work. 

I say re-read what Aron and RG said, with the thought that these guys have done this dozens (probably hundreds) of times and are experts.
cheers

[Zben3129]

Hi
depending on the current capacity of that 12V supply, there are a few options, assuming that you want to ignore Aron's advice and do it the difficult and inefficient way instead of the simple and effective way.

Assuming that you've got 100mA or thereabout of current capacity, try putting a 22 ohm 5 W resistor to the +, in series with a 100 ohm 5W pot to ground.  Connect your rig to the junction.  Then adjust the pot until you get about 9V at the junction. As RG said, the voltage will sag with extra load and rise with less load.  But it will kind of work. 

I say re-read what Aron and RG said, with the thought that these guys have done this dozens (probably hundreds) of times and are experts.
cheers

After the first experiment or two I pretty much knew this was a very touchy/near useless way to control voltage, I just wanted to figure out how/why/what are the ways around it. Its kinda like I started a useless puzzle. Even if I finish it it won't be anything useful. But I will have the satisfaction of completing it and also the knowledge to not do it again and/or how to make it more useful next time.

More of a learning kinda deal than a practical at this point.

Zach

[George Giblet]

The easiest way to drop a small amount of voltage is to add a few diodes in series, 4 diodes gives 2.8V drop.  If you want the no load voltage to measure close to the loaded voltage add something like a 10k resistor to ground - it's a little academic though in practice.

Another solution is to use a vbe multiplier tuned to the voltage drop you want (again you need to put a dummy load to get the no load voltage to be right.)

A third solution is to use an voltage divide followed by an emitter follower:

http://www.technology.niagarac.on.ca/courses/elnc1231/Files/Unit_4.pdf
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/emitfol.html

A fourth solution is to use a replace the resistor to ground R2 with a zener (third diagram on second page).

The emitter followers also need a dummy load to make the no load voltage match the loaded.

All the above methods have poorer load regulation to a regulator but they are good enough in practice and simple/cheap.

[DougH]

You can also wire up a power MOSFET to drop voltage. I do that to drop 50-100v in my amps (see my schematics for reference). But with circuits like this,  by the time you do that you may as well do it the right way with a regulator. IC regulators are super simple to use, the only "out of the ordinary" thing is they require heat sinking. There's no reason not to use them for this kind of stuff.